Ta có:
\(ab=c\Rightarrow\frac{a}{1}=\frac{c}{b}\)
\(ac=9b\Rightarrow\frac{c}{b}=\frac{9}{a}\)
\(\Rightarrow\frac{a}{1}=\frac{9}{a}\left(=\frac{c}{b}\right)\)
\(\Rightarrow a^2=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-3\end{matrix}\right.\)
TH1:\(a=3\)
\(\Rightarrow\frac{c}{b}=3\Rightarrow c=3b\)
\(\Rightarrow bc=b3b=3b^2=4a=4.3=12\)
\(\Rightarrow b^2=4\Rightarrow\left[{}\begin{matrix}b=2\\b=-2\end{matrix}\right.\)
Nếu \(b=2\)\(\Rightarrow c=6\)
Nếu \(b=-2\Rightarrow c=-6\)
TH2: \(a=-3\)
\(\Rightarrow\frac{c}{b}=-3\Rightarrow c=-3b\)
\(\Rightarrow bc=b\left(-3b\right)=-3b^2=4a=4.\left(-3\right)=-12\)
\(\Rightarrow b^2=4\Rightarrow\left[{}\begin{matrix}b=2\\b=-2\end{matrix}\right.\)
Nếu \(b=2\Rightarrow c=-6\)
Nếu \(b=-2\Rightarrow c=6\)
Vậy \(\left(a;b;c\right)\in\left\{\left(3;2;6\right);\left(3;-2;-6\right);\left(-3;2;-6\right);\left(-3;-2;6\right)\right\}\) thỏa mãn đề bài
ABC cân tại A, góc A = 500:
ABH=
ACH.